Answer:
[tex]$ \frac{\sqrt{3} - 1}{2\sqrt{2}} $[/tex]
[tex]$ \frac{-(\sqrt{3} + 1)}{2\sqrt{2}} $[/tex]
[tex]$ - \frac{\sqrt{3} - 1}{\sqrt{3} + 1} $[/tex]
Step-by-step explanation:
Given [tex]$ \frac{11 \pi}{12} = \frac{3 \pi}{4} + \frac{\pi}{6} $[/tex]
(A) [tex]$ sin(\frac{11\pi}{12}) = sin (\frac{3 \pi}{4} + \frac{\pi}{6}) $[/tex]
We know that Sin(A + B) = SinA cosB + cosAsinB
Substituting in the above formula we get:
[tex]$ sin (\frac{3\pi}{4} + \frac{\pi}{6}) = \frac{1}{\sqrt{2}} . \frac{\sqrt{3}}{2} + \frac{-1}{\sqrt{2}}. \frac{1}{2} $[/tex]
[tex]$ \implies \frac{1}{\sqrt{2}} (\frac{\sqrt{3} - 1}{2}) = \frac{\sqrt{3} - 1}{2\sqrt{2}}[/tex]
(B) Cos(A + B) = CosAcosB - SinASinB
[tex]$ cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6}}) $[/tex]
[tex]$ \implies \frac{-1}{\sqrt{2}}. \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} . \frac{1}{2} $[/tex]
[tex]$ \implies cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6}) $[/tex]
[tex]$ = \frac{-(\sqrt{3} + 1)}{2\sqrt{2}}[/tex]
(C) Tan(A + B) = [tex]$ \frac{Sin(A +B)}{Cos(A + B)} $[/tex]
From the above obtained values this can be calculated and the value is [tex]$ - \frac{\sqrt{3} - 1}{\sqrt{3} + 1} $[/tex].
